Examples of factoring polynomials. How to Factor a Quadratic Trinomial: Formula Trinomial Equation

Expanding polynomials to obtain a product can sometimes seem confusing. But it's not that difficult if you understand the process step by step. The article describes in detail how to factor a quadratic trinomial.

Many people do not understand how to factor a square trinomial and why this is done. At first it may seem like a futile exercise. But in mathematics nothing is done for nothing. The transformation is necessary to simplify the expression and ease of calculation.

A polynomial of the form – ax²+bx+c, called a quadratic trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say it differently: how to expand a quadratic equation.

Interesting! A polynomial is called a square because of its largest degree, the square. And a trinomial - because of the 3 components.

Some other types of polynomials:

• linear binomial (6x+8);
• cubic quadrinomial (x³+4x²-2x+9).

Factoring a quadratic trinomial

First, the expression is equal to zero, then you need to find the values ​​of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. You need to know its formula by heart: D=b²-4ac.

If the result D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated using the formula.

If, when calculating the discriminant, the result is zero, you can use any of the formulas. In practice, the formula is simply shortened: -b / 2a.

The formulas for different discriminant values ​​are different.

If D is positive:

If D is zero:

Online calculators

There is an online calculator on the Internet. It can be used to perform factorization. Some resources provide the opportunity to view the solution step by step. Such services help to better understand the topic, but you need to try to understand it well.

Useful video: Factoring a quadratic trinomial

Examples

We suggest looking at simple examples of how to factor a quadratic equation.

Example 1

This clearly shows that the result is two x's because D is positive. They need to be substituted into the formula. If the roots turn out to be negative, the sign in the formula changes to the opposite.

We know the formula for factoring a quadratic trinomial: a(x-x1)(x-x2). We put the values ​​in brackets: (x+3)(x+2/3). There is no number before a term in a power. This means that there is one there, it goes down.

Example 2

This example clearly shows how to solve an equation that has one root.

We substitute the resulting value:

Example 3

Given: 5x²+3x+7

First, let's calculate the discriminant, as in previous cases.

D=9-4*5*7=9-140= -131.

The discriminant is negative, which means there are no roots.

After receiving the result, you should open the brackets and check the result. The original trinomial should appear.

Alternative solution

Some people were never able to make friends with the discriminator. There is another way to factorize a quadratic trinomial. For convenience, the method is shown with an example.

Given: x²+3x-10

We know that we should get 2 brackets: (_)(_). When the expression looks like this: x²+bx+c, at the beginning of each bracket we put x: (x_)(x_). The remaining two numbers are the product that gives "c", i.e. in this case -10. The only way to find out what numbers these are is by selection. The substituted numbers must correspond to the remaining term.

For example, multiplying the following numbers gives -10:

• -1, 10;
• -10, 1;
• -5, 2;
• -2, 5.

1. (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
2. (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
3. (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
4. (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.

This means that the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).

Important! You should be careful not to confuse the signs.

Expansion of a complex trinomial

If “a” is greater than one, difficulties begin. But everything is not as difficult as it seems.

To factorize, you first need to see if anything can be factored out.

For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:

3(x²+3x-10). The result is the already well-known trinomial. The answer looks like this: 3(x-2)(x+5)

How to decompose if the term that is in the square is negative? In this case, the number -1 is taken out of brackets. For example: -x²-10x-8. The expression will then look like this:

The scheme differs little from the previous one. There are just a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets that need to be filled in (_)(_). In the 2nd bracket is written x, and in the 1st what is left. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.

The number 3 is given by the numbers:

• -1, -3;
• -3, -1;
• 3, 1;
• 1, 3.

We solve equations by substituting these numbers. The last option is suitable. This means that the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).

Other cases

It is not always possible to convert an expression. With the second method, solving the equation is not required. But the possibility of transforming terms into a product is checked only through the discriminant.

It is worth practicing solving quadratic equations so that when using the formulas there are no difficulties.

Useful video: factoring a trinomial

Conclusion

You can use it in any way. But it’s better to practice both until they become automatic. Also, learning how to solve quadratic equations well and factor polynomials is necessary for those who are planning to connect their lives with mathematics. All the following mathematical topics are built on this.

In contact with

This online calculator is designed to factorize a function.

For example, factorize: x 2 /3-3x+12. Let's write it as x^2/3-3*x+12. You can also use this service, where all calculations are saved in Word format.

For example, decompose into terms. Let's write it as (1-x^2)/(x^3+x) . To see the progress of the solution, click Show steps. If you need to get the result in Word format, use this service.

Note: the number "pi" (π) is written as pi; square root as sqrt , for example sqrt(3) , tangent tg is written tan . To view the answer, see Alternative.

1. If a simple expression is given, for example, 8*d+12*c*d, then factoring the expression means representing the expression in the form of factors. To do this, you need to find common factors. Let's write this expression as: 4*d*(2+3*c) .
2. Present the product in the form of two binomials: x 2 + 21yz + 7xz + 3xy. Here you already need to find several common factors: x(x+7z) + 3y(x + 7z). We take out (x+7z) and get: (x+7z)(x + 3y) .

see also Division of polynomials with a corner (all steps of division with a column are shown)

Useful when studying the rules of factorization will be abbreviated multiplication formulas, with the help of which it will be clear how to open brackets with a square:

1. (a+b) 2 = (a+b)(a+b) = a 2 +2ab+b 2
2. (a-b) 2 = (a-b)(a-b) = a 2 -2ab+b 2
3. (a+b)(a-b) = a 2 - b 2
4. a 3 +b 3 = (a+b)(a 2 -ab+b 2)
5. a 3 -b 3 = (a-b)(a 2 +ab+b 2)
6. (a+b) 3 = (a+b)(a+b) 2 = a 3 +3a 2 b + 3ab 2 +b 3
7. (a-b) 3 = (a-b)(a-b) 2 = a 3 -3a 2 b + 3ab 2 -b 3

Factorization Methods

1. Using abbreviated multiplication formulas.
2. Finding a common factor.

Factoring a quadratic trinomial may be useful when solving inequalities from problem C3 or problem with parameter C5. Also, many B13 word problems will be solved much faster if you know Vieta’s theorem.

This theorem, of course, can be considered from the perspective of the 8th grade, in which it is taught for the first time. But our task is to prepare well for the Unified State Exam and learn to solve exam tasks as efficiently as possible. Therefore, this lesson considers an approach slightly different from the school one.

Formula for the roots of the equation using Vieta’s theorem Many people know (or at least have seen):

$$x_1+x_2 = -\frac(b)(a), \quad x_1 x_2 = \frac(c)(a),$$

where `a, b` and `c` are the coefficients of the quadratic trinomial `ax^2+bx+c`.

To learn how to easily use the theorem, let's understand where it comes from (this will actually make it easier to remember).

Let us have the equation `ax^2+ bx+ c = 0`. For further convenience, divide it by `a` and get `x^2+\frac(b)(a) x + \frac(c)(a) = 0`. This equation is called a reduced quadratic equation.

Important lesson idea: any quadratic polynomial that has roots can be expanded into parentheses. Let's assume that ours can be represented as `x^2+\frac(b)(a) x + \frac(c)(a) = (x + k)(x+l)`, where `k` and ` l` - some constants.

Let's see how the brackets open:

$$(x + k)(x+l) = x^2 + kx+ lx+kl = x^2 +(k+l)x+kl.$$

Thus, `k+l = \frac(b)(a), kl = \frac(c)(a)`.

This is slightly different from the classical interpretation Vieta's theorem- in it we look for the roots of the equation. I propose to look for terms for bracket decomposition- this way you don’t need to remember about the minus from the formula (meaning `x_1+x_2 = -\frac(b)(a)`). It is enough to select two such numbers, the sum of which is equal to the average coefficient, and the product is equal to the free term.

If we need a solution to the equation, then it is obvious: the roots `x=-k` or `x=-l` (since in these cases one of the brackets will be zero, which means the entire expression will be zero).

I'll show you the algorithm as an example: How to expand a quadratic polynomial into brackets.

Example one. Algorithm for factoring a quadratic trinomial

The path we have is a quadrant trinomial `x^2+5x+4`.

It is reduced (the coefficient of `x^2` is equal to one). He has roots. (To be sure, you can estimate the discriminant and make sure that it is greater than zero.)

Further steps (you need to learn them by completing all training tasks):

1. Complete the following entry: $$x^2+5x+4=(x \ldots)(x \ldots).$$ Instead of dots, leave free space, we will add suitable numbers and signs there.
2. Consider all possible options for decomposing the number `4` into the product of two numbers. We get pairs of “candidates” for the roots of the equation: `2, 2` and `1, 4`.
3. Figure out which pair you can get the average coefficient from. Obviously it's `1, 4`.
4. Write $$x^2+5x+4=(x \quad 4)(x \quad 1)$$.
5. The next step is to place signs in front of the inserted numbers.

   How to understand and forever remember what signs should appear before the numbers in brackets? Try opening them (brackets). The coefficient before `x` to the first power will be `(± 4 ± 1)` (we don’t know the signs yet - we need to choose), and it should be equal to `5`. Obviously, there will be two pluses $$x^2+5x+4=(x + 4)(x + 1)$$.

   Perform this operation several times (hello, training tasks!) and you will never have any more problems with this.

If you need to solve the equation `x^2+5x+4`, then now solving it will not be difficult. Its roots are `-4, -1`.

Example two. Factorization of a quadratic trinomial with coefficients of different signs

Let us need to solve the equation `x^2-x-2=0`. Offhand, the discriminant is positive.

We follow the algorithm.

1. $$x^2-x-2=(x \ldots) (x \ldots).$$
2. There is only one decomposition of two into integer factors: `2 · 1`.
3. We skip the point - there is nothing to choose from.
4. $$x^2-x-2=(x \quad 2) (x \quad 1).$$
5. The product of our numbers is negative (`-2` is the free term), which means that one of them will be negative and the other will be positive.
   Since their sum is equal to `-1` (the coefficient of `x`), then `2` will be negative (the intuitive explanation is that two is the larger of the two numbers, it will “pull” more strongly in the negative direction). We get $$x^2-x-2=(x - 2) (x + 1).$$

Third example. Factoring a quadratic trinomial

The equation is `x^2+5x -84 = 0`.

1. $$x+ 5x-84=(x \ldots) (x \ldots).$$
2. Factorization of 84 into integer factors: `4·21, 6·14, 12·7, 2·42`.
3. Since we need the difference (or sum) of the numbers to be 5, the pair `7, 12` is suitable.
4. $$x+ 5x-84=(x\quad 12) (x\quad 7).$$
5. $$x+ 5x-84=(x + 12) (x - 7).$$

Hope, expansion of this quadratic trinomial into brackets It's clear.

If you need a solution to an equation, here it is: `12, -7`.

Training tasks

I bring to your attention a few examples that are easy are solved using Vieta's theorem.(Examples taken from the magazine "Mathematics", 2002.)

1. `x^2+x-2=0`
2. `x^2-x-2=0`
3. `x^2+x-6=0`
4. `x^2-x-6=0`
5. `x^2+x-12=0`
6. `x^2-x-12=0`
7. `x^2+x-20=0`
8. `x^2-x-20=0`
9. `x^2+x-42=0`
10. `x^2-x-42=0`
11. `x^2+x-56=0`
12. `x^2-x-56=0`
13. `x^2+x-72=0`
14. `x^2-x-72=0`
15. `x^2+x-110=0`
16. `x^2-x-110=0`
17. `x^2+x-420=0`
18. `x^2-x-420=0`

A couple of years after the article was written, a collection of 150 tasks for expanding a quadratic polynomial using Vieta’s theorem appeared.

Like and ask questions in the comments!

In this lesson we will learn how to factor quadratic trinomials into linear factors. To do this, we need to remember Vieta’s theorem and its converse. This skill will help us quickly and conveniently expand quadratic trinomials into linear factors, and will also simplify the reduction of fractions consisting of expressions.

So let's go back to the quadratic equation, where .

What we have on the left side is called a quadratic trinomial.

The theorem is true: If are the roots of a quadratic trinomial, then the identity holds

Where is the leading coefficient, are the roots of the equation.

So, we have a quadratic equation - a quadratic trinomial, where the roots of the quadratic equation are also called the roots of the quadratic trinomial. Therefore, if we have the roots of a square trinomial, then this trinomial can be decomposed into linear factors.

Proof:

The proof of this fact is carried out using Vieta’s theorem, which we discussed in previous lessons.

Let's remember what Vieta's theorem tells us:

If are the roots of a quadratic trinomial for which , then .

The following statement follows from this theorem:

We see that, according to Vieta’s theorem, i.e., by substituting these values ​​into the formula above, we obtain the following expression

Q.E.D.

Recall that we proved the theorem that if are the roots of a square trinomial, then the expansion is valid.

Now let's remember an example of a quadratic equation, to which we selected roots using Vieta's theorem. From this fact we can obtain the following equality thanks to the proven theorem:

Now let's check the correctness of this fact by simply opening the brackets:

We see that we factored correctly, and any trinomial, if it has roots, can be factorized according to this theorem into linear factors according to the formula

However, let's check whether such factorization is possible for any equation:

Take, for example, the equation . First, let's check the discriminant sign

And we remember that in order to fulfill the theorem we learned, D must be greater than 0, so in this case, factorization according to the theorem we learned is impossible.

Therefore, we formulate a new theorem: if a square trinomial has no roots, then it cannot be decomposed into linear factors.

So, we have looked at Vieta’s theorem, the possibility of decomposing a quadratic trinomial into linear factors, and now we will solve several problems.

Task No. 1

In this group we will actually solve the problem inverse to the one posed. We had an equation, and we found its roots by factoring it. Here we will do the opposite. Let's say we have the roots of a quadratic equation

The inverse problem is this: write a quadratic equation using its roots.

There are 2 ways to solve this problem.

Since are the roots of the equation, then is a quadratic equation whose roots are given numbers. Now let's open the brackets and check:

This was the first way in which we created a quadratic equation with given roots, which does not have any other roots, since any quadratic equation has at most two roots.

This method involves the use of the inverse Vieta theorem.

If are the roots of the equation, then they satisfy the condition that .

For the reduced quadratic equation , , i.e. in this case, and .

Thus, we have created a quadratic equation that has the given roots.

Task No. 2

It is necessary to reduce the fraction.

We have a trinomial in the numerator and a trinomial in the denominator, and the trinomials may or may not be factorized. If both the numerator and the denominator are factored, then among them there may be equal factors that can be reduced.

First of all, you need to factor the numerator.

First, you need to check whether this equation can be factorized, let’s find the discriminant. Since , the sign depends on the product (must be less than 0), in this example, i.e. the given equation has roots.

To solve, we use Vieta’s theorem:

In this case, since we are dealing with roots, it will be quite difficult to simply select the roots. But we see that the coefficients are balanced, that is, if we assume that , and substitute this value into the equation, we get the following system: , i.e. 5-5=0. Thus, we have selected one of the roots of this quadratic equation.

We will look for the second root by substituting what is already known into the system of equations, for example, , i.e. .

Thus, we have found both roots of the quadratic equation and can substitute their values ​​into the original equation to factor it:

Let's remember the original problem, we needed to reduce the fraction .

Let's try to solve the problem by substituting .

It is necessary not to forget that in this case the denominator cannot be equal to 0, i.e. , .

If these conditions are met, then we have reduced the original fraction to the form .

Problem No. 3 (task with a parameter)

At what values ​​of the parameter is the sum of the roots of the quadratic equation

If the roots of this equation exist, then , question: when.

Online calculator.
Isolating the square of a binomial and factoring a square trinomial.

Those. the problems boil down to finding the numbers \(p, q\) and \(n, m\)

The program not only gives the answer to the problem, but also displays the solution process.

This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

If you are not familiar with the rules for entering a quadratic trinomial, we recommend that you familiarize yourself with them.

Rules for entering a quadratic polynomial

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.

Numbers can be entered as whole or fractional numbers.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.

Rules for entering decimal fractions.
In decimal fractions, the fractional part can be separated from the whole part by either a period or a comma.
For example, you can enter decimal fractions like this: 2.5x - 3.5x^2

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &
Input: 3&1/3 - 5&6/5x +1/7x^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) x + \frac(1)(7)x^2\)

When entering an expression you can use parentheses. In this case, when solving, the introduced expression is first simplified.
For example: 1/2(x-1)(x+1)-(5x-10&1/2)

Example of a detailed solution

Isolating the square of a binomial.$$ ax^2+bx+c \rightarrow a(x+p)^2+q $$ $$2x^2+2x-4 = $$ $$2x^2 +2 \cdot 2 \cdot\left( \frac(1)(2) \right)\cdot x+2 \cdot \left(\frac(1)(2) \right)^2-\frac(9)(2) = $$ $$2\left (x^2 + 2 \cdot\left(\frac(1)(2) \right)\cdot x + \left(\frac(1)(2) \right)^2 \right)-\frac(9 )(2) = $$ $$2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Answer:$$2x^2+2x-4 = 2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Factorization.$$ ax^2+bx+c \rightarrow a(x+n)(x+m) $$ $$2x^2+2x-4 = $$
$$ 2\left(x^2+x-2 \right) = $$
$$ 2 \left(x^2+2x-1x-1 \cdot 2 \right) = $$ $$ 2 \left(x \left(x +2 \right) -1 \left(x +2 \right ) \right) = $$ $$ 2 \left(x -1 \right) \left(x +2 \right) $$ Answer:$$2x^2+2x-4 = 2 \left(x -1 \right) \left(x +2 \right) $$

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A little theory.

Isolating the square of a binomial from a square trinomial

If the square trinomial ax 2 +bx+c is represented as a(x+p) 2 +q, where p and q are real numbers, then we say that from square trinomial, the square of the binomial is highlighted.

From the trinomial 2x 2 +12x+14 we extract the square of the binomial.

\(2x^2+12x+14 = 2(x^2+6x+7) \)

To do this, imagine 6x as a product of 2*3*x, and then add and subtract 3 2. We get:
$$ 2(x^2+2 \cdot 3 \cdot x + 3^2-3^2+7) = 2((x+3)^2-3^2+7) = $$ $$ = 2 ((x+3)^2-2) = 2(x+3)^2-4 $$

That. We extract the square binomial from the square trinomial, and showed that:
$$ 2x^2+12x+14 = 2(x+3)^2-4 $$

Factoring a quadratic trinomial

If the square trinomial ax 2 +bx+c is represented in the form a(x+n)(x+m), where n and m are real numbers, then the operation is said to have been performed factorization of a quadratic trinomial.

Let us show with an example how this transformation is done.

Let's factor the quadratic trinomial 2x 2 +4x-6.

Let us take the coefficient a out of brackets, i.e. 2:
\(2x^2+4x-6 = 2(x^2+2x-3) \)

Let's transform the expression in brackets.
To do this, imagine 2x as the difference 3x-1x, and -3 as -1*3. We get:
$$ = 2(x^2+3 \cdot x -1 \cdot x -1 \cdot 3) = 2(x(x+3)-1 \cdot (x+3)) = $$
$$ = 2(x-1)(x+3) $$

That. We factored the quadratic trinomial, and showed that:
$$ 2x^2+4x-6 = 2(x-1)(x+3) $$

Note that factoring a quadratic trinomial is possible only if the quadratic equation corresponding to this trinomial has roots.
Those. in our case, it is possible to factor the trinomial 2x 2 +4x-6 if the quadratic equation 2x 2 +4x-6 =0 has roots. In the process of factorization, we established that the equation 2x 2 + 4x-6 = 0 has two roots 1 and -3, because with these values, the equation 2(x-1)(x+3)=0 turns into a true equality.